Question: Simplify the following expression: $y = \dfrac{5x^2+19x- 30}{5x - 6}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(-30)} &=& -150 \\ {a} + {b} &=& &=& {19} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-150$ and add them together. Remember, since $-150$ is negative, one of the factors must be negative. The factors that add up to ${19}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-6}$ and ${b}$ is ${25}$ $ \begin{eqnarray} {ab} &=& ({-6})({25}) &=& -150 \\ {a} + {b} &=& {-6} + {25} &=& 19 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 {-6}x) + ({25}x {-30}) $ Factor out the common factors: $ x(5x - 6) + 5(5x - 6)$ Now factor out $(5x - 6)$ $ (5x - 6)(x + 5)$ The original expression can therefore be written: $ \dfrac{(5x - 6)(x + 5)}{5x - 6}$ We are dividing by $5x - 6$ , so $5x - 6 \neq 0$ Therefore, $x \neq \frac{6}{5}$ This leaves us with $x + 5; x \neq \frac{6}{5}$.